如圖①,△BCD內(nèi)接于直角梯形
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,A
1D∥A
2A
3,A
1A
2⊥A
2A
3,A
1D=10,A
1A
2=8,沿△BCD三邊將△A
1BD、△A
2BC、△A
3CD翻折上去,恰好形成一個三棱錐ABCD,如圖②.
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(1)求證:AB⊥CD;
(2)求直線BD和平面ACD所成的角的正切值;
(3)求四面體
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的體積。
(1)詳見解析;(2)
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; (3)
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試題分析:(1)平面圖中因為A
1D∥A
2A
3,A
1A
2⊥A
2A
3,所以
,立體圖中不變,即
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,可證得
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,就可證出AB⊥CD。(2)由(1)知AB⊥平面ACD.,所以AD即為BD在面ACD內(nèi)的射影,所以∠BDA即為所求。在直角三角形中利用三角函數(shù)可求其正切值。(3)由(1)知
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,所以可以選以面ADC為底面,以AB為高求其體積。
試題解析:(1)證明:∵在直角梯形A
1A
2A
3D中,A
1B⊥A
1D,A
2B⊥A
2C,
∴在三棱錐ABCD中,AB⊥AD,AB⊥AC.
∵AC∩AD=A,∴AB⊥平面ACD.
∵CD?平面ACD,∴AB⊥CD.
(2)解:由(1)知AB⊥平面ACD,
∴AD為BD在平面ACD內(nèi)的射影,
∠BDA是直線BD和平面ACD所成的角.
依題意,在直角梯形A
1A
2A
3D中,
A
1D=A
3D=10,A
1B=A
2B=4,
∴在三棱錐ABCD中,AD=10,AB=4.
在Rt△ABD中,tan ∠BDA=
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=
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=
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.
∴直線BD和平面ACD所成的角的正切值為

.
(3)由(2)得:
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,直線m
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①
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∥
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⊥m;②
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⊥
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∥m;③
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∥m
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⊥
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;④
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⊥m
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∥
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其中正確命題序號是
.
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